How to check if something is perfect square or not.

Question

While finding eigenvalues of a particular matrix, I end up with the following: $\sqrt{p^{2\alpha}-4p^{\alpha}+8p^{\alpha-1}+4}$, where $p$ is an odd prime and $\alpha \geq 1$. The next step is to check if this eigenvalue is an integer or not. How do I go about checking the integrality of such an expression?

Answer 1

Edited to add: Please stop upvoting me! Piquito's is the answer that deserves your votes!

If $p\ge 11$:**
That eigenvalue is not an integer. This is because $p^{2\alpha}-4p^\alpha+4=(p^\alpha-2)^2$ is a perfect square, and the next perfect square is $(p^\alpha-1)^2=p^{2\alpha}-2p^\alpha+1$, which is strictly greater than $p^{2\alpha}-4p^\alpha+8p^{\alpha-1}+4$ if $p\ge 11$ (because rearranging terms, this is equivalent to $2p^{\alpha-1}(p-8)>3$, which is true for all $\alpha\ge 1$).

So $p^{2\alpha}-4p^\alpha+8p^{\alpha-1}+4$ lies strictly between two consecutive perfect squares, and therefore is not itself a square.

If $\alpha=1$:
The expression inside the square root reduces to $p^2-4p+12=(p-2)^2+8$; the only two perfect squares that differ by $8$ are $1$ and $9$, so $(p-2)^2=1$, giving $p=3$. So $\alpha=1,p=3$ gives the eigenvalue $3$.

Answer 2

Since the radical is equal to $(p^{\alpha}-2)^2+8p^{\alpha-1}$ we have the equation in three unkowns $(p,\alpha,k)$ $$p^{2\alpha}-4p^{\alpha}+8p^{\alpha-1}+4=(p^{\alpha}-2+k)^2=p^{2\alpha}-4p^{\alpha}+2kp^{\alpha}-4k+k^2+4\\8p^{\alpha-1}=2kp^{\alpha}-4k+k^2$$ so $k$ is even and $$2p^{\alpha-1}=k_1(p^{\alpha}-1) $$ which implies $k_1=2$ or $k_1=p$ so we have respectively $$p^{\alpha-1}=p^{\alpha}-1\\2p^{\alpha-1}=p(p^{\alpha}-1)$$ The only solution is given by $(3^1-2)^2+8\times3^0=3^2$; in other words the only integer eigenvalue is $3$.